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dimaraw [331]
4 years ago
11

The following frequency table summarizes yesterday's orders at Sizzlin' Skillets.

Mathematics
2 answers:
Tcecarenko [31]4 years ago
4 0

Answer:

0.28

Step-by-step explanation

Getting it wrong on Khan Academy hah

Alexxandr [17]4 years ago
4 0

Answer:

0.28

Step-by-step explanation:

You might be interested in
Does anyone know the answer to this
Marysya12 [62]
Hello!

We can write this as a proportion below.

\frac{blueprint}{real} = \frac{0.5}{4} = \frac{18}{x}

As you can see, 4 is eight times 0.5. Therefore, we can multiply eighteen by eight to get our answer.

18(8)=144

Just to check, we can cross multiply and divide.

72/0.5=144

Therefore, our answer is 144 feet.

I hope this helps!
3 0
3 years ago
10 - 8y = 10 - 8y<br> Is it a solution
Aleks [24]
It’s all reall numbers
5 0
3 years ago
Write the first four terms in the following sequences. A(n+1)=A(n)−5 for n≥1 and A(1)=9 .
luda_lava [24]

Answer:

Step-by-step explanation:

A(1) = 9

A(n+1) = A(n) - 5

n = 1 ;     A(1+1) = A(1) - 5

             A(2) =  9 - 5

             A(2) = 4

n = 2 ;     A(2+1) = A(2) - 5

             A(3) =  4 - 5

             A(3) = 1

n = 3 ;     A(3+1) = A(3) - 5

             A(4) =  1 - 5

             A(4) = -4

n = 4 ;     A(4+1) = A(4) - 5

             A(5) =  -4 - 5

             A(5) = -9

First four terms are: 9 , 4 , 1 , -4 , - 9

4 0
3 years ago
Convert please
dezoksy [38]
<h3>Answer:</h3>
  1. 3,8 kg
  2. 750 g
  3. 7,47 g
<h3>Step-by-step explanation:</h3><h3>3800 g => kg</h3>

(3800 ÷ 1000) kg

= 3,8 kg

<h3>¾ kg => g</h3>

(3 ÷ 4)kg × 1000

= 0.75 × 1000

= 750 g

<h3>7470 g => kg</h3>

(7470 ÷ 1000) kg

= 7,47 kg

4 0
3 years ago
What should a size of a sample be in order to have a margin of error of 4% for a 90% confidence level from a population with a s
vagabundo [1.1K]

Answer:

We need a sample size of 2,071,800 or higher.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.9}{2} = 0.05

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.05 = 0.95, so z = 1.645

Now, we find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

In this problem:

We need a sample size of n or higher, when M = 0.04, \sigma = 35. So

M = z*\frac{\sigma}{\sqrt{n}}

0.04 = 1.645*\frac{35}{\sqrt{n}}

0.04\sqrt{n} = 1.645*35

0.04\sqrt{n} = 57.575

\sqrt{n} = 1439.375

\sqrt{n}^{2} = (1439.375)^{2}

n = 2,071,800

We need a sample size of 2,071,800 or higher.

6 0
3 years ago
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